Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 19

Write the formal Maclaurin series as ${\sum }_{n\ge 0}{a}_n{x}^n$, where $a_n=f^{(n)}(0)/n!$. If the successive nonzero coefficient ratios have a limit, the ratio test givesso the series converges absolutely for $|x|<R$, whereIf that ratio does not exist, the Cauchy-Hadamard theorem gives the general formula $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. The radius of a Maclaurin series from successive derivatives concerns convergence of the series; equality to the original function additionally requires analyticity or a vanishing Taylor remainder. Mere differentiability, or even smoothness, does not ensure that equality.

For the specified logarithm,The absolute successive-term ratio tends to $|x|/2$, so the interval of absolute convergence is $(-2,2)$. At $x=2$ the nonconstant terms are the alternating harmonic series, which converges conditionally to $\ln 2$, giving the full sum $\ln 4$. At $x=-2$ all nonconstant terms are $-1/n$, so the series diverges to $-\infty$; the original logarithm is also undefined there. Thus the full real convergence interval is $(-2,2]$.

Multiply the Maclaurin series $e^{-x}=1-x+x^2/2+O(x^3)$ by the binomial series $(1+x{)}^{-1/2}=1-x/2+3x^2/8+O(x^3)$. The constant, linear and quadratic coefficients are therefore

There is an indexing defect in the original PDF: the term at $n=1$ is undefined because $\ln 1=0$. Thus the series as literally indexed is not defined. The natural finite-index repair is to start at $n=2$, and that repaired series still diverges.

Indeed, for $n\ge 2$ the terms are positive andThey eventually increase at a geometric rate and do not tend to zero. Since the terms of a convergent series tend to zero, no choice of a finite replacement for the missing first term could make the tail converge.

Use the bound $|\sin n|\le 1$:The comparison p-series has exponent $2>1$ and converges. Hence, by the comparison test for series, the original series is absolutely convergent. No alternating-sign or monotonicity assumption about $\sin n$ is needed.

Index the terms from $n=0$:Its absolute successive-term ratio tends to zero for every fixed $x$, so it converges everywhere. Divide by $x$ temporarily and use the exponential series divided by a shifted index with $k=3$:ThereforeAt $x=0$ both the original series and this formula are zero, so the temporary division loses no value. The termwise integration is justified by uniform convergence of the exponential series on the finite integration interval.